Answer : The correct answer for the Theoretical Yield is 48.93 g of product .
Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .
Following are the steps to find theoretical yield .
Step 1) : Write a balanced reaction between Al and Cl₂ .
2 Al + 3 Cl₂→ 2 AlCl₃
Step 2: To find amount of product (AlCl₃) formed by Al .
Following are the sub steps to calculate amount of AlCl₃ formed :
a) To calculate mole of Al :
Given : Mass of Al = 35.5 g
Mole can be calculate by following formula :
![Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}](https://tex.z-dn.net/?f=%20Mole%20%3D%20%5Cfrac%7Bgiven%20mass%20%28g%29%7D%7Batomic%20mass%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20)
![Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}](https://tex.z-dn.net/?f=%20Mole%20%3D%20%5Cfrac%7B35.5%20g%20%7D%7B26.9%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20)
Mole = 1.32 mol
b) To find mole ratio of AlCl₃ : Al
Mole ratio is calculated from balanced reaction .
Mole of Al in balanced reaction = 2
Mole of AlCl₃ in balanced reaction = 2.
Hence mole ratio of AlC; l₃ : Al = 2:2
c) To find mole of AlCl₃ formed :
![Mole of AlCl_3 = Mole of Al * Mole ratio](https://tex.z-dn.net/?f=%20Mole%20of%20AlCl_3%20%3D%20Mole%20of%20Al%20%2A%20Mole%20ratio%20%20)
![Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}](https://tex.z-dn.net/?f=%20Mole%20of%20AlCl_3%20%3D%201.32%20mol%20of%20Al%20%2A%20%5Cfrac%7B2%7D%7B2%7D%20%20)
Mole of AlCl₃ = 1.32 mol
d) To find mass of AlCl₃
Molar mass of AlCl₃ = ![133.34 \frac{g}{mol}](https://tex.z-dn.net/?f=%20133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20)
Mass of AlCl3 can be calculated using mole formula as:
![1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}](https://tex.z-dn.net/?f=%201.32%20mol%20of%20AlCl_3%20%3D%20%5Cfrac%7B%20mass%20%28g%29%7D%7B133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20)
Multiplying both side by ![133.34 \frac{g}{mol}](https://tex.z-dn.net/?f=%20133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20)
![1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}](https://tex.z-dn.net/?f=%201.32%20mole%20%20%2A%20133.34%5Cfrac%7Bg%7D%7Bmol%7D%20%3D%20%5Cfrac%7Bmass%20%28g%29%7D%7B133.34%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%2A133.34%20%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20)
Mass of AlCl₃ = 176.00 g
Hence mass of AlCl₃ produced by Al is 176.00 g
Step 3) To find mass of product (AlCl₃) formed by Cl₂ :
Same steps will be followed to calculate mass of AlCl₃
a) Find mole of Cl₂
![Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}](https://tex.z-dn.net/?f=%20Mole%20of%20Cl_2%20%3D%20%5Cfrac%7B39.0%20g%7D%7B70.9%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20)
Mole of Cl₂ = 0.55 mol
b) Mole ratio of Cl₂ : AlCl₃
Mole of Cl₂ in balanced reaction = 3
Mole of AlCl₃ in balanced reaction = 2
Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3
c) To find mole of AlCl₃
![Mole of AlCl_3 = mole of Cl_2 * mole ratio](https://tex.z-dn.net/?f=%20Mole%20of%20AlCl_3%20%3D%20mole%20of%20Cl_2%20%2A%20mole%20ratio%20%20)
![Mole of AlCl_3 = 0.55 mole * \frac{2}{3}](https://tex.z-dn.net/?f=%20Mole%20of%20AlCl_3%20%3D%200.55%20%20mole%20%20%2A%20%5Cfrac%7B2%7D%7B3%7D%20%20)
Mole of AlCl3 = 0.367 mol
d) To find mass of AlCl₃ :
![0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}](https://tex.z-dn.net/?f=%200.367%20mol%20of%20AlCl_3%20%3D%20%5Cfrac%7Bmass%20%28g%29%20%7D%7B133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20)
Multiplying both side by
![133.34 \frac{g}{mol}](https://tex.z-dn.net/?f=%20133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20)
![0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}](https://tex.z-dn.net/?f=%200.367%20mol%20of%20AlCl_3%20%2A%20133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20%3D%20%5Cfrac%7Bmass%28g%29%7D%7B133.34%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%20%20%2A%20133.34%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20)
Mass of AlCl₃ = 48.93 g
Hence mass of AlCl₃ produced by Cl₂ = 48.93 g
Step 4) To identify limiting reagent and theoretical yield :
Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .
The product AlCl₃ formed by Al = 176.00 g
The product AlCl₃ formed by Cl₂ = 48.93 g
Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .